A Eulogy Of Lipschitz Maps

A Lipschitz map (/function) is one that does not extend distances by more than a pre-assigned factor: $f: X \longrightarrow Y$ is Lipschitz if there exists an $L \in \mathbb{R}$ such that

$$ \forall x, \ \ \forall y \ \ \ d(f(x),f(y)) \leq L d(x,y) \ .$$

The definition makes sense as long as a distance is defined on the spaces. This makes Lipschitz maps highly versatile. Almost every space you deal with daily is either a priori a metric space (a set plus a distance function) or can be made one by endowing it with a distance. (Check for instance the word metric on a group that builds metric spaces out of groups, opening the doors to the beautiful topic of Geometric Group Theory.)

We will, nonetheless, limit ourselves to studying (some of) the properties of Lipschitz maps between Euclidean spaces. Properties?! Look at the definition again, what else could we expect of a Lipschitz map? What else could that condition possibly impose on $f$?

One of the most surprising facts about Lipschitz maps is the following:

Radamacher’s Theorem: If $f: \mathbb{R}^n \longrightarrow \mathbb{R}^m$ is Lipschitz, then it is differentiable a.e. (i.e is differentiable everywhere on $\mathbb{R}^n$ except maybe on a set of Lebesgue measure zero.)

Not surprised? Let’s look at the simplest case where $f: \mathbb{R}^1 \longrightarrow \mathbb{R}^1$ and all we know about $f$ is this:

Theorem (McShane): Assume $A$ is a subset of a metric space $X$ — so, it is a metric space itself — and $f:A \longrightarrow Y$ is Lipschitz. Then there exists a map $F: X \longrightarrow Y$ such that $F(a)=f(a)$ for all $a \in A$, $F$ is Lipschitz with the Lipschitz constant (the $L$ in the definition) the same as that of $f$.

This property shows that the differentiability result above holds if $\mathbb{R}^n$ is replaced with an open $\Omega \subseteq \mathbb{R}^n$.

By Radamacher’s theorem, a Lipschitz map $f: \mathbb{R}^n \supseteq \Omega \longrightarrow \mathbb{R}^n$ has derivative a.e on $\Omega$. Thus, its Jacobian is defined a.e. which leads us to another theorem.

provided that $u\geq 0$ or one of the integrands is integrable. Measurability of the integrands is part of the assertion of the theorem as is the fact that the integrability of one of the integrands implies that of the other.

There are many versions of this theorem and many different proofs in different generalities. A nice and complete proof (of a more general case) can be found in [2].

In the realm of integration, the area formula and co-area formula hold for Lipschitz maps. (Theorem 3.23 in [1])

Note how much easier it is to show that a map is Lipschitz than is to verify for instance its differentiability and (local) integrability of the derivative. Some form of integrability/boundedness is required of the derivative of a function to satisfy the change of variables formula. This observation tells us that Lipscitz maps are in fact “nicer” than differentiable maps. Their derivatives are smoother and tamer, in some sense. The following theorem is a testimony to this notion.

Observe that $C^1$ maps are locally Lipschitz by the Mean Value Theorem because on compact sets the derivative is bounded. So, the closest cousin to Lipschitz functions are $C^1$ functions. In fact, many proofs about Lipschitz maps can be reduced to verifications for $C^1$ cases.

Moving to the next property on our list, the Luzin N property: A Lipschitz map from $\mathbb{R}^n$ to $\mathbb{R}^n$ that takes measure zero sets to measure zero sets. This interestingly implies that any measurable set is mapped to a measurable set. Of course, for a differentiable map (not $C^1$) this does not have to hold — another reason why Lipschitz maps are more than being a.e. differentiable.

Since the definition of a Lipschitz map is through the distance function, of course Hausdorff measures must come into the picture somewhere:

Theorem: If $f:X \longrightarrow Y$ is Lipschitz between metric spaces with Lipschitz bound $L$, then for any $E \subseteq X$, and any $ s $,

$$ \mathcal{H}^s (f(E)) \leq L^s \mathcal{H}^s (E) \ ,$$

which in particular gives the Luzin N property above.

Finally, I finish with a fact that my advisor (Piotr Hajlasz) and I verified recently:

Theorem: Let $f:\mathbb{R}^m \longrightarrow \mathbb{R}^n$ be Lipschitz. If the derivative of $f$ at a (single) point has rank $k$, then $Df$ will have rank at least $k$ on a set of positive $m$-measure.

Again, note that the property is an elementary fact if the map is $C^1$.

The V-shaped $y=|x|$ is not that broken after all!

If you know of some other facts about Lipschitz maps that you find cool, please write in the comments!

About Behnam Esmayli

I started PhD in Mathematics at Pitt in Fall 2015. I have come to grow a passion for metric spaces -- a set and a distance function that satisfies the triangle inequality -- simple and beautiful! These spaces when equipped with other structures, such as a measure, becomes extremely fun to play with!

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